Để ;(x + 1).(x - 3) < 0 thì ta có 2 trường hợp

Th1 : \(\hept{\begin{cases}x+1< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>3\end{cases}\left(loai\right)}}\)

Th2 : \(\hept{\begin{cases}x+1>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow}-1< x< 3}\)

Ta có : 5x³ - 45x

= 5x(x² - 9)

= 5x(x - 3)(x + 3)

b ) Ta có : 3x² - 7x - 6 

= 3x² - 9x + 2x - 6 

= 3x (x - 3) + 2(x - 3)

= (x - 3)(3x + 2)

a,

\(\left(x+1\right)\left(x-3\right)< 0\)

\(\Rightarrow x+1\text{ và }x-3\text{ khác dấu và }x+1\ne0,x-3\ne0\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-3< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\Rightarrow1< x< 3\\\left\{{}\begin{matrix}x+1< 0\\x-3>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>3\end{matrix}\right.\Rightarrow\text{mâu thuẫn}\end{matrix}\right.\)

Vậy \(1< x< 3\) thì \(\left(x+1\right)\left(x-3\right)< 0\)

b,

\(\dfrac{x+1}{x-4}>0\)

\(\Rightarrow x+1\text{ và }x-4\text{ cùng dấu và }x+1\ne0,x-4\ne0\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-4>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x>4\end{matrix}\right.\Rightarrow x>4\\\left\{{}\begin{matrix}x+1< 0\\x-4< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x< 4\end{matrix}\right.\Rightarrow x< -1\end{matrix}\right.\)

Vậy khi \(x>4\) hoặc \(x< -1\) thì \(\dfrac{x+1}{x-4}>0\)

\(\left(x+1\right)\left(x-3\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-3>0\Rightarrow x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1< x< 3\)

\(\dfrac{x+1}{x-4}>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-4>0\Rightarrow x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-4< 0\Rightarrow x< 4\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x>-1;x< 4\)

a) Để (x+1)(x-3) <0

Thì ta có 2 TH:

+x+1>0 ; x-3<0

+x+1<0 ; x-3>0

Th1:x+1>0 ; x-3<0

+x+1>0=>x>-1

+x-3<0=>x<3

=>3>x>-1(TM)

Th2:x+1<0 ; x-3>0

+x+1<0=>x<-1

+x-3>0=>x>3

=>-1>x>3 (Loại vì vô lí)

Vậy (x+1).(x-3)<0 <=>3>x>-1

b)Để x+1/x-4 >0

=>x+1>x-4 và x+1; x-4 đều cùng lớn hơn hoặc bé hơn 0

Tự làm tiếp nhá

x=3

b,Dat an 2x^2-3x-1=a la dc

a, \(4^x-10.2^x+16=0\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Đặt \(2^x=t\Rightarrow t^2-10t+16=0\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

b. Đặt \(2x^2-3x-1=t\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Leftrightarrow t^2-3t-28=0\Leftrightarrow\orbr{\begin{cases}t=7\\t=-4\end{cases}}\)

Thế vào rồi giải tiếp em nhé.

camon ạ

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

1. Rút gọn biểu thức :

\(M=4.\left(2-3x\right)-\left|2x-3\right|\) (*)

- Xét 2 TH :

+ Trường hợp 1 : \(\left|2x-3\right|=\left(2x-3\right)\) thì (*) trở thành :

\(M=4.\left(2-3x\right)-\left(2x-3\right)\)

\(\Rightarrow M=8-12x-2x+3\)

\(\Rightarrow M=-14x+11\)

+ Trường hợp 2 : \(\left|2x-3\right|=\left(3-2x\right)\) thì (*) trở thành :

\(M=4.\left(2-3x\right)-\left(3-2x\right)\)

\(\Rightarrow M=8-12x-3+2x\)

\(\Rightarrow M=-10x+5\)

Cái bài giải phương trình ở lớp 8 mới học nhé bạn.

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)

mà A\(\le0\)

​​\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\)​ phải bằng 0 đê thỏa mãn điều kiện

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy....

b;c)I hệt câu a nên làm tương tự nhá

d)

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)

B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)

Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)

Vậy....

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|\ge0\\\left|y+\dfrac{4}{3}\right|\ge0\\\left|z+\dfrac{7}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\Rightarrow x=-\dfrac{9}{2}\\\left|y+\dfrac{4}{3}\right|=0\Rightarrow y=-\dfrac{4}{3}\\\left|z+\dfrac{7}{2}\right|=0\Rightarrow z=-\dfrac{7}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\\left|y-\dfrac{2}{5}\right|\ge0\\\left|z+\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}\right|=0\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|\ge0\\ \left|y+\dfrac{1980}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\Rightarrow x=-\dfrac{19}{5}\\ \left|y+\dfrac{1980}{1975}\right|=0\Rightarrow y=-\dfrac{1980}{1975}\\\left|z-2004\right|=0\Rightarrow z=2004\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\ \left|y-\dfrac{1}{5}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{1}{5}\right|=0\Rightarrow y=\dfrac{1}{5}\\\left|x+y+z\right|=0\Rightarrow z+-\dfrac{11}{20}=0\Rightarrow z=\dfrac{11}{20}\end{matrix}\right.\)

Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\Rightarrow A\ge0\)

Mà ĐK đề là \(A\le0\)

\(\Rightarrow A=0\)

\(\left[{}\begin{matrix}\left|x+\dfrac{3}{4}=0\right|\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}=0\right|\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}=0\right|\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

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